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As shown in the figure, a block of mass ...

As shown in the figure, a block of mass `sqrt3` kg is kept on a horizontal rough surface of coefficient of friction `1/(3sqrt3)` The critical force to be applied on the vertical surface as shown at an angle `60^@` with horizontal such that it does not move, will be 3x. The value of x will be
`[g=10m//s^2,sin60^@=sqrt3/2,cos60^@=1/2]`

Text Solution

Verified by Experts

The correct Answer is:
`3.33`

`F cos60^(@)=muN` or `(F)/(2)=(1)/(3sqrt(3))N`........(i)
and `N=sin60^(@)+sqrt(3)g`........(ii)
From equation (i) & (ii)
From equation (i) & (ii)
`(F)/(2)=(1)/(3sqrt(3))((Fsqrt(3))/(2)+sqrt(3)g)`
`impliesF=g=10N=3x`
So, `x=(10)/(3)=3.33`
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